Related Topics:

Linear Combinations, Linear Independence

Second‐order differential equations involve the second derivative of the unknown function (and, quite possibly, the first derivative as well) but no derivatives of higher order. For nearly every second‐order equation encountered in practice, the general solution will contain two arbitrary constants, so a second‐order IVP must include two initial conditions.

Given two functions y1( x) and y2( x), any expression of the form

where c1 and c2 are constants, is called a linear combination of y1 and y2. For example, if y1 = exand y2 = x2, then

are all particular linear combinations of y1 and y2. So the idea of a linear combination of two functions is this: Multiply the functions by whatever constants you wish; then add the products.

Note that this fits the form of a linear combination of sin x and cos x,

by taking c1 = cos 1 and c2 = sin 1.

Example 3: Can the function y = x3 be written as a linear combination of the functions y1 = x and y2 = x2?

If the answer were yes, then there would be constants c1 and c2 such that the equation

holds true for all values of x. Letting x = 1 in this equation gives

and letting x = −1 gives

Adding these last two equations gives 0 = 2 c2, so c2 = 0. And since c2 = 0, c1 must equal 1. Thus, the general linear combination (*) reduces to

which clearly does not hold for all values of x. Therefore, it is not possible to write y = x3 as a linear combination of y1 = x and y2 = x2.

One more definition: Two functions y1 and y2 are said to be linearly independent if neither function is a constant multiple of the other. For example, the functions y1 = x3 and y2 = 5 x3 are not linearly independent (they're linearly dependent), since y2 is clearly a constant multiple of y1. Checking that two functions are dependent is easy; checking that they're independent takes a little more work.

If they weren't, then y1 would be a constant multiple of y2; that is, the equation

would hold for some constant c and for all x. But substituting x = π/2, for example, yields the absurd statement 1 = 0. Therefore, the above equation cannot be true: y1 = sin x is not a constant multiple of y2 = cos x; thus, these functions are indeed linearly independent.

Example 5: Are the functions y1 = exand y2 = x linearly independent?

If they weren't, then y1 would be a constant multiple of y2; that is, the equation

would hold for some constant c and for all x. But this cannot happen, since substituting x = 0, for example, yields the absurd statement 1 = 0. Therefore, y1 = exis not a constant multiple of y2 = x; these two functions are linearly independent.

Example 6: Are the functions y1 = xexand y2 = exlinearly independent?

A hasty conclusion might be to say no because y1 is a multiple of y2. But y1 is not a constant multiple of y2, so these functions truly are independent. (You may find it instructive to prove they're independent by the same kind of argument used in the previous two examples.)